Question

Find the domain and the range of the real function, $f\left(x\right)=\frac{3}{(2-x^2)}$.

Answer 1

We have, $f\left(x\right)=\frac{3}{(2-x^2)}$.

Clearly, f(x) is defined for all real values of x except those for which

$2-x^2=0$,i.e., $x=\pm \sqrt{2}$

$\therefore$ dom (f) =$R-\{-\sqrt{2},\sqrt{2}\}.$

Let $y=f\left(x\right)$. Then,

$y=\frac{3}{(2-x^2)}\Rightarrow 2y-x^{2}y=3\Rightarrow x^{2}y=2y-3$

$\Rightarrow x^2=\frac{2y-3}{y}\Rightarrow x=\pm \sqrt{\frac{2y-3}{y}}\dots \dots \left(i\right)$

It is clear from (i) that x will take real values only when $\frac{2y-3}{y}\ge 0$.

Now, $\frac{2y-3}{y}\ge 0\iff (2y-3\le 0\text{and}y<0)$ or $(2y-3\ge 0\text{and}y>0)$

$\iff (y\le \frac{3}{2}\text{and}y<0)$ or $(y\ge \frac{3}{2}\text{and}y>0)$

$\iff (y<0)$ or $(y\ge \frac{3}{2})$

$\iff yϵ(-\infty ,0)$ or $yϵ[\frac{3}{2},\infty )$

$\iff yϵ(-\infty ,0)\cup [\frac{3}{2},\infty )$

$\therefore$ range (f) $=(-\infty ,0)\cup [\frac{3}{2},\infty )$

Hence, $\text{dom (f)}=R-\{-\sqrt{2},\sqrt{2}\}$ and range (f) $=(-\infty ,0)\cup [\frac{3}{2},\infty )$